Question

If work done in stretching a wire by $1mm$ is $2J$, the work necessary for stretching another wire of same material, but with double the radius and half the length by $1mm$ in joule is -

• A. $1/4$
• B. $4$
• C. $8$
• D. $16$

Answer 1

The correct option is D $16$

$\begin{array}{c}\text{Work done}=U=\frac{1}{2}\times \text{stress}x\text{strain}x\text{velume}\\ \begin{array}{cc}U& =\frac{1}{2}xY\times (\text{strain}{)}^2\times \text{volume}\\ U& =\frac{1}{2}\times Y\times \frac{(\Delta l{)}^2}{l^2}\times Al\end{array}\end{array}$

$\begin{array}{c}=\frac{1}{2}xYx(\Delta l{)}^2\times \frac{A}{l}=\frac{1}{2}xyx(\Delta \rho {)}^{2}x\pi \frac{R^2}{l}\\ U=\left(\frac{1}{2}xy\pi x\right(\Delta l{)}^{2}x)\times \frac{R^2}{l}\end{array}$

$\begin{array}{c}2J=[\frac{1}{2}\times y\pi x{\left({10}^{-3}\right)}^2]\times \frac{R^2}{l}\\ U=[\frac{1}{2}\times y\pi x{\left({10}^{-3}\right)}^2]\times \frac{(2R{)}^2}{\left(\frac{2}{2}\right)}\\ \text{from equation (1) and (2)}\\ U=16J\end{array}$